[proofplan]
We prove disconnectedness by exhibiting a separation of the disjoint union. Choose one nonempty summand $X_i$ and take its canonical image $A$ inside the coproduct space. The coproduct topology makes each canonical summand open, and the complement of one summand is the union of all the other summands, hence also open. Since another nonempty summand $X_j$ exists with $j \neq i$, both $A$ and its complement are nonempty, giving a separation.
[/proofplan]
[step:Name the coproduct space and isolate one canonical summand]
Let $Y := \bigsqcup_{k \in I} X_k$ denote the topological disjoint union equipped with the coproduct topology. For each $k \in I$, let $\iota_k: X_k \to Y$ denote the canonical injection into the $k$-th summand.
Choose distinct indices $i,j \in I$ with $X_i \neq \varnothing$ and $X_j \neq \varnothing$. Define the subset $A \subset Y$ by
\begin{align*}
A := \iota_i(X_i).
\end{align*}
Since $X_i \neq \varnothing$, choose $x_i \in X_i$. Then $\iota_i(x_i) \in A$, so $A \neq \varnothing$.
[/step]
[step:Show the chosen summand is open and has open complement]
By the definition of the coproduct topology, a subset $O \subset Y$ is open if and only if $\iota_k^{-1}(O) \in \tau_k$ for every $k \in I$.
For the set $A = \iota_i(X_i)$, the canonical images of distinct summands in a disjoint union are disjoint. Therefore $\iota_i^{-1}(A) = X_i$. For every $k \in I$ with $k \neq i$, $\iota_k^{-1}(A) = \varnothing$. Since $X_i \in \tau_i$ and $\varnothing \in \tau_k$ for every $k \in I$, the coproduct-topology criterion implies that $A$ is open in $Y$.
Now define $C := Y \setminus A$. Again using disjointness of the canonical summands, $\iota_i^{-1}(C) = \varnothing$. For every $k \in I$ with $k \neq i$, $\iota_k^{-1}(C) = X_k$. Since $\varnothing \in \tau_i$ and $X_k \in \tau_k$ for every $k \in I$, the coproduct-topology criterion implies that $C$ is open in $Y$.
[guided]
The point of using the canonical injections is that the summands of a disjoint union are not treated as overlapping subsets of some ambient space; each summand enters $Y$ through its own map. For every $k \in I$, we have a canonical injection $\iota_k: X_k \to Y$.
The coproduct topology is defined so that a subset $O \subset Y$ is open exactly when every inverse image $\iota_k^{-1}(O)$ is open in the corresponding space $(X_k,\tau_k)$. We apply this criterion first to $A := \iota_i(X_i)$. Because different canonical summands in a disjoint union are disjoint, the inverse image of $A$ under $\iota_i$ is all of $X_i$, so $\iota_i^{-1}(A) = X_i$. If $k \neq i$, then the $k$-th summand does not meet the $i$-th summand, so $\iota_k^{-1}(A) = \varnothing$. The set $X_i$ is open in its own topology $\tau_i$, and $\varnothing$ is open in every topology. Hence all inverse images of $A$ under the canonical injections are open, so $A$ is open in $Y$.
Now consider the complement $C := Y \setminus A$. Its inverse images are complementary to those just computed. In the $i$-th summand, no point remains after removing $A$, so $\iota_i^{-1}(C) = \varnothing$. For each $k \neq i$, the whole $k$-th summand remains inside $C$, so $\iota_k^{-1}(C) = X_k$. Again these inverse images are open in their respective spaces. Therefore $C$ is open in $Y$ by the definition of the coproduct topology.
[/guided]
[/step]
[step:Use the second nonempty summand to obtain a separation]
Since $X_j \neq \varnothing$, choose $x_j \in X_j$. Because $j \neq i$, the point $\iota_j(x_j)$ belongs to the summand $\iota_j(X_j)$, which is disjoint from $A = \iota_i(X_i)$. Hence $\iota_j(x_j) \in C$, so $C \neq \varnothing$.
The sets $A$ and $C$ are open in $Y$, nonempty, disjoint, and satisfy $A \cup C = Y$. Thus $A$ and $C$ form a separation of $Y$. Therefore the topological disjoint union $Y = \bigsqcup_{k \in I} X_k$ is disconnected.
[/step]
